Question: Find the slope and y-intercept of the line that is ${\text{perpendicular}}$ to $\enspace {y = -\dfrac{1}{2}x + 2}\enspace$ and passes through the point ${(2, 8)}$. ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$ ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$
Solution: Lines are considered perpendicular if their slopes are negative reciprocals of each other. The slope of the blue line is ${-\dfrac{1}{2}}$ , and its negative reciprocal is ${2}$ Thus, the equation of our perpendicular line will be of the form $\enspace {y = 2x + b}\enspace$ We can plug our point, $(2, 8)$ , into this equation to solve for ${b}$ , the y-intercept. $8 = {2}(2) + {b}$ $8 = 4 + {b}$ $8 - 4 = {b} = 4$ The equation of the perpendicular line is $\enspace {y = 2x + 4}\enspace$. ${m = 2, \enspace b = 4}$